Finding side lengths using trigonometric ratios
Pythagoras' theorem enabled us to find the lengths of an unknown side of a right-angled triangle given the two other sides. The trigonometric ratios, sine, cosine and tangent provide the means to calculate the lengths of the sides of a right-angled triangle.
If we know one angle (other than the right angle) and one side length in a right-angled triangle, then we can then find any other side in the same triangle.
The most important part of the process is to correctly identify the ratio that relates the angle and the sides we are interested in. The following examples show the process:
Worked examples
Example 1
Determine the value of $b$b in the triangle below:
Think: In this right-angled triangle we have:
- a $25^\circ$25° angle
- a hypotenuse of length $12.6$12.6 units
- the side opposite $25^\circ$25° of unknown length
The two sides are the opposite side and the hypotenuse, so we can use the sine ratio.
Do:
| $\sin\theta$sinθ | $=$= | $\frac{\text{Opposite }}{\text{Hypotenuse }}$Opposite Hypotenuse |
Set up the $\sin$sin ratio |
| $\sin25^\circ$sin25° | $=$= | $\frac{b}{12.6}$b12.6 |
Substituting values for the known angle and sides into the ratio |
| $b$b | $=$= | $12.6\times\sin25^\circ$12.6×sin25° |
Rearranging the equation to solve for $b$b |
| $=$= | $5.32$5.32 ($2$2 d.p.) |
Using our calculator to evaluate $b$b, rounding to two decimal places |
The first step is the most important. We need to identify which ratio will help us relate the angle and sides. Once we have this, we just need to solve the equation correctly.
Reflect: Note that the unknown was in the numerator and the equation was solved by multiplying throughout by the denominator of the ratio we set up.
Example 2
Determine the value of $c$c in the triangle below:
Think: In this right-angled triangle we have:
- a $36^\circ$36° angle
- the hypotenuse of unknown length
- the adjacent side of length $4.8$4.8 units
The two sides are the adjacent side and the hypotenuse, so we can use the cosine ratio.
Do:
| $\cos\theta$cosθ | $=$= | $\frac{\text{Adjacent }}{\text{Hypotenuse }}$Adjacent Hypotenuse |
Set up the $\cos$cos ratio |
| $\cos36^\circ$cos36° | $=$= | $\frac{4.8}{c}$4.8c |
Substituting values into the ratio |
| $c\times\cos36^\circ$c×cos36° | $=$= | $4.8$4.8 |
Multiplying both sides of the equation by $c$c |
| $c$c | $=$= | $\frac{4.8}{\cos36^\circ}$4.8cos36° |
Dividing by $\cos36$cos36 to make $c$c the subject |
| $=$= | $5.93$5.93 ($2$2 d.p.) |
Using our calculator to evaluate $c$c, rounding to two decimal places |
Reflect: Note that the unknown was in the denominator and the equation required a couple of extra steps to solve compared to the previous example.
When the missing side is on the top of the trig ratio (numerator) we multiply by the sin, cos or tan of the angle.
When the missing side is on the bottom of the trig ratio (denominator) we divide by the sin, cos or tan of the angle.
Practice questions
Question 1
Find the value of $f$f, correct to two decimal places.
An inverted right triangle with its base on top. The angle on the left vertex of the base is $47^\circ$47°. Adjacent to this angle is the base of the triangle labeled as "$f$f m," indicating its unknown length. The angle on the right vertex of the base is a right angle as indicated by a small square. Opposite the right angle is the hypotenuse labeled "$8$8 m," indicating its length.
Question 2
Find the value of $h$h, correct to two decimal places.
Finding angles using trigonometric ratios
The trigonometric ratios, sine, cosine and tangent also enable us to calculate the internal angles of a right-angled triangle.
If we know two side lengths, then we can then find the unknown angles of a right-angled triangle.
We can use the following process to find an angle:
- Label the sides as opposite, adjacent or hypotenuse with respect to the angle we want to find.
- Identify the trigonometric ratio of this angle which involves the known lengths of two sides.
- Write and solve the equation for this ratio of the unknown angle.
When we set up the trigonometric ratio to find an angle we end up with an equation containing an expression like $\sin x$sinx. To isolate the angle ($x$x in this case) from the trigonometric ratio ($\sin$sin in this case) we need to use what is called the inverse trigonometric ratios.
Consider the triangle below:
We can see that the $\sin$sin ratio of theta is $0.5$0.5. So we can form the equation $\sin\theta=0.5$sinθ=0.5. This equation is saying that if we take the $\sin$sin of the angle $\theta$θ, the ratio of the length of the opposite side to the hypotenuse is $0.5$0.5. Or in other words, the hypotenuse is twice the length of the opposite side with respect to the angle $\theta$θ. When we take the $\sin$sin of an angle we get the ratio of the sides. The opposite of this is taking the inverse $\sin$sin, designated $\sin^{-1}$sin−1, of the ratio to give the angle.
We 'undo' the trigonometric ratio of an angle by applying an inverse trigonometric ratio to both sides of the equation as follows:
| $\sin\theta$sinθ | $=$= | $x$x |
| $\sin^{-1}\left(\sin\theta\right)$sin−1(sinθ) | $=$= | $\sin^{-1}\left(x\right)$sin−1(x) |
| $\theta$θ | $=$= | $\sin^{-1}\left(x\right)$sin−1(x) |
The three inverse trigonometric ratios are $\sin^{-1}$sin−1, $\cos^{-1}$cos−1, and $\tan^{-1}$tan−1 and they are defined so that:
If $\sin\theta=x$sinθ=x, then $\theta=\sin^{-1}(x)$θ=sin−1(x)
If $\cos\theta=y$cosθ=y, then $\theta=\cos^{-1}(y)$θ=cos−1(y)
If $\tan\theta=z$tanθ=z, then $\theta=\tan^{-1}(z)$θ=tan−1(z)
In the case of the above example, if $\sin\theta=0.5$sinθ=0.5, then $\theta=\sin^{-1}(0.5)$θ=sin−1(0.5).
Finding the inverse with a calculator
We can use these inverse trigonometric functions by pressing the appropriate button on a calculator. The inverse functions are usually recalled by using the using shift and the $\sin$sin, $\cos$cos and $\tan$tan buttons.
If you enter $\sin^{-1}(0.5)$sin−1(0.5) in your calculator, you get the result $30^\circ$30°. This tells us that an angle of $30^\circ$30° has a sine ratio of $0.5$0.5.
Worked examples
Example 3
Find the angle $\theta$θ in the triangle below to the nearest degree.
Think: We first want to identify the sides in relation to the angle, then determine the appropriate trigonometric ratio to use.
Do:
- Label the sides as opposite, adjacent or hypotenuse with respect to $\theta$θ.
- Identify the ratio that relates the opposite side and the hypotenuse. For this angle, it will be $\sin$sin.
- Write and solve the equation for $\sin\theta$sinθ.
| $\sin\theta$sinθ | $=$= | $\frac{\text{Opposite }}{\text{Hypotenuse }}$Opposite Hypotenuse |
write the rule |
| $\sin\theta$sinθ | $=$= | $\frac{5}{8}$58 |
fill in the known side lengths |
| $\theta$θ | $=$= | $\sin^{-1}\left(\frac{5}{8}\right)$sin−1(58) |
find the missing angle by doing the inverse(opposite) of $\sin$sin on your calculator |
| $\theta$θ | $=$= | $39^\circ$39° |
use a calculator to evaluate $\theta$θ to the nearest degree |
Example 4
Find the angle $\theta$θ in the triangle below to the nearest degree.
Think: In this triangle, we know the lengths of the opposite and adjacent sides. So the ratio we will use is $\tan$tan.
Do:
| $\tan\theta$tanθ | $=$= | $\frac{\text{Opposite }}{\text{Adjacent }}$Opposite Adjacent |
write the rule |
| $\tan\theta$tanθ | $=$= | $\frac{14.77}{12.24}$14.7712.24 |
fill in the known side lengths |
| $\theta$θ | $=$= | $\tan^{-1}\left(\frac{14.77}{12.24}\right)$tan−1(14.7712.24) |
find the missing angle by doing the inverse(opposite) of $\tan$tan on your calculator |
| $\theta$θ | $=$= | $50^\circ$50° |
use a calculator to evaluate $\theta$θ to the nearest degree |
Practice questions
question 3
Find the value of $x$x to the nearest degree.
A right-angled triangle with vertices labeled A, B and C. Vertex A is at the top, B at the bottom right, and C at the bottom left. A small square at vertex A indicates that it is a right angle. Side interval(BC), which is the side opposite vertex A, is the hypotenuse and is marked with a length of 25. The angle located at vertex B is labelled x. Side interval(AB), descending from the right angle at vertex A to vertex B, is marked with a length of 7, and is adjacent to the angle x. Side interval(AC) is opposite the angle x.
question 4
Use the tangent ratio to find the size of the angle marked $x$x, correct to the nearest degree.
